# Getting zero as solution to the 1D wave equation

Your solution $A_n$ isn't zero for *every* $n$. Looking at your initial conditions you might expect there to be only one non-zero coefficient since the IC already have the form of a sine wave. When $n=1$ you get $\frac 0 0$ so you have to be extra careful. Evaluating your solution for $A_n$ as a limit:
$$\lim_{n\rightarrow 1}\frac{2\sin n\pi}{\pi(n^2-1)}=-1$$
I suspect there is an error in the integral because the answer should be +1. Alternatively you could immediately see from
$$\sin\left(\frac \pi l x\right)=\sum_{n=1}^\infty A_n \sin\left(\frac {n\pi} l x\right)$$
that $A_n=\cases{1&$n=1$\\0 &$n\neq 1$}$

Wolframalpha is likely the source of confusion here: the initial condition is the term with $n=1$, i.e. $A_1 = 1$, whereas all others are $A_n=0$. The equation for the initial condition is one equation with the infinite number of unknowns in the rhs - no surprise that an automatic routine gives an irrelevant answer there.